Find the general solution to each equation, leaving answers in radian form: i) sqrt2sin2theta + 1 = 0 & ii) cos4x + cos2x = 0

Question

The answer to both are:

i) (n x Pi)/2 + (-1)^n x ((5 x Pi)/8) 

ii) n x Pi +- Pi/6 or n x Pi +- Pi/2

I’d greatly appreciate of you explained what ‘n’ is and explain how to answer the steps you make along the way in order to answer the questions and get to the answers stated before, please and thank you!

Answer 1

If your first equation is

i) √2 sin(2θ) + 1 = 0, then we can solve it as following:

sin(2θ)=-1/√2

2θ =-π/4+2nπ, n=0, ±1,±2, ±3,...

θ =-π/8+nπ, n=0, ±1,±2, ±3,...

n is an integer, which shows basically the number of periods.

 

To get a correct and precise answer, your parentheses to make sure the expression in the RHS is correct.

 

ii) cos4x + cos2x = 0  

[

Use cos(2u)=cos²u-sin²u=2cos²u-1. Then cos(4x)=2cos²(2x)-1].

2cos²(2x)-1+cos2x=0

 

Substitution cos(2x)=y

2y²+y-1=0

Solve this equation (y=1/2amd y=-1) and then solve

a) cos(2x)=-1/2and b) cos(2x)=-1.

 

 

Answer 2

let x=θ to simplify the work

i) √2sin2x+1=0

 

√2sin2x+1=0

√2sin2x=-1

sin2x=-1/√2

sin2x=-√2/2

sin equals -√2/2 at 225º (5∏/4) and at 315º (7∏/4)

therefore 2x=5∏/4 and x=5∏/8 (answer)

also 2x=7∏/4 and x=7∏/8 (answer)

 

ii) cos4x+cos2x=0

 

cos4x+cos2x=0

let y=2x

therefore cos4x=cos2y and cos2x=cosy

cos2y+cosy=0

cos2y=2cos²y-1 (trig identity)

2cos²y-1+cosy=0

2cos²y+cosy-=1=0

factor

(2cosy-1)(cosy+1)=0

2cosy-1=0 or cosy+1=0

2cosy-1=0

2cosy=1

cosy=1/2

cosy equals 1/2 at ∏/3 and at 5∏/3

but y=2x

therefore 2x=∏/3 and x=∏/6 (answer)

also 2x=5∏/3 and x=5∏/6 (answer)

now work with cosy+1=0

cosy=-1

cosy equals -1 at ∏

but y=2x

therefore 2x=∏ and x=∏/2 (answer)

if cos(2x)=y for example, then 2x=y+360n and x=y/2 + 180n for n=0,±1,±2,... to check for more values